Find the normalization transformation that maps a window whose lower left corner is at (1,1) and upper right corner is at (3, 5) onto a viewport that is the entire normalized device screen.

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{ - \frac{1}{4}}&{\frac{1}{4}}\\ 1&1&1 \end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\\ 0&0&1 \end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{ - \frac{1}{4}}&{\frac{1}{4}}\\ 1&1&1 \end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\\ 0&{\frac{1}{4}}&{\frac{1}{4}}\\ 1&0&0 \end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\\ 1&0&0 \end{array}} \right)\)

Explore topic-wise MCQs in Computer Graphics.

This section includes 3 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Graphics knowledge and support exam preparation. Choose a topic below to get started.

1.	A 4 x 4 DFT matrix is given by :\(\frac{1}{2}\left[ {\begin{array}{{20}{c}} {\begin{array}{{20}{c}} 1&1\\ 1&x \end{array}}&{\begin{array}{{20}{c}} 1&1\\ { - 1}&y \end{array}}\\ {\begin{array}{{20}{c}} 1&{ - 1}\\ 1&{ - j} \end{array}}&{\begin{array}{*{20}{c}} 1&{ - 1}\\ { - 1}&j \end{array}} \end{array}} \right]\)(j2 = -1)Where values of x and y are_____ , ______ respectively.
A.	1, -1
B.	-1, 1
C.	-j, j
D.	j, -j
Answer» E.

Discussion

2.	Given that a 22-inch monitor with an aspect ratio of 16 : 9 has a monitor resolution of 1920 X 1080, what is the width of the monitor?
A.	8.53 inches
B.	19.17 inches
C.	10.79 inches
D.	22 inches
Answer» C. 10.79 inches

Discussion

3.	Find the normalization transformation that maps a window whose lower left corner is at (1,1) and upper right corner is at (3, 5) onto a viewport that is the entire normalized device screen.
A.	\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\\ 0&0&1 \end{array}} \right)\)
B.	\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{ - \frac{1}{4}}&{\frac{1}{4}}\\ 1&1&1 \end{array}} \right)\)
C.	\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\\ 0&{\frac{1}{4}}&{\frac{1}{4}}\\ 1&0&0 \end{array}} \right)\)
D.	\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\\ 1&0&0 \end{array}} \right)\)
Answer» B. \(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{\frac{1}{2}}\\ 0&{ - \frac{1}{4}}&{\frac{1}{4}}\\ 1&1&1 \end{array}} \right)\)

Discussion

Explore topic-wise MCQs in Computer Graphics.

Given that a 22-inch monitor with an aspect ratio of 16 : 9 has a monitor resolution of 1920 X 1080, what is the width of the monitor?

Find the normalization transformation that maps a window whose lower left corner is at (1,1) and upper right corner is at (3, 5) onto a viewport that is the entire normalized device screen.